Is a right-continuous function on a compact space even "uniformly right-continuous"?

Let $E$ be a normed $\mathbb R$-vector space, $I\subseteq\mathbb R$ be nonempty, $O\subseteq I$ be open and $x:I\to\mathbb R$. Assume $x$ is right-continuous on $O$ and $$x(t-):=\lim_{s\to t-}x(s)$$ exists for all $t\in O$. Let $$\Delta x(t):=x(t)-x(t-)\;\;\;\text{for }t\ge0.$$

Let $K\subseteq O$ be compact. Are we able to show that $x$ is uniformly right-continuous on $K$? Likewise, are we able to deduce a uniformity of the left-limits on $K$? In particular, I would like to show that for all $\varepsilon>0$, there is a $\delta>0$ such that $$\left\|x(u)-x(s)-\Delta x(t)\right\|_E<\varepsilon\tag1$$ for all $t\in K$, $s\in(t-\delta,\delta)$ and $u\in[t,t+\delta)$.

$(1)$ is easy to show for a fixed $t\in O$ and $\delta$ depending on $t$.

I've tried to proceed as in the proof of uniform continuity of continuous functions on compact spaces:

Since $x$ has left-limits on $O$, there is a $(\delta^-_t)_{t\in O}\subseteq(0,\infty)$ with $$\forall t\in O:\forall s\in I\cap(t-\delta^-_t,t):\left\|x(s)-x(t-)\right\|_E<\frac\varepsilon3\tag2.$$ Analogously, since $x$ is right-continuous on $O$, there is a $(\delta^+_t)_{t\in O}\subseteq(0,\infty)$ with $$\forall t\in O:\forall u\in I\cap[t,t+\delta^+_t):\left\|x(t)-x(u)\right\|_E<\frac\varepsilon3\tag3.$$

Now let's try to show to obtain the uniformity of the left-limits: $\bigcup_{t\in O}\left(t-\frac{\delta^-_t}2,t\right)$ is an open cover of $K$ (this is the reason why I've introduced the open neighborhood $O$ of $K$, since I guess we couldn't proceed without $O$ (i.e. if $O=K$)) and hence $$K\subseteq\bigcup_{i=1}^n\left(t_i-\frac{\delta^-_{t_i}}2,t_i\right)\tag4$$ for some $n\in\mathbb N$ and $t_1,\ldots,t_n\in O$. Let $$\delta:=\frac12\min_{1\le i\le n}\delta^-{t_i}.$$

Now we would need to show that if $t\in K$ and $s\in I\cap(t-\delta,t)$, then $\left\|x(s)-x(t-)\right\|_E<\varepsilon$: Since $t\in K$, there is an $i\in\{1,\ldots,n\}$ with $t\in\left(t_i-\frac{\delta^-_{t_i}}2,t_i\right)$. This clearly implies $s\in(t_i-\delta^-_{t_i},t_i)$. However, I struggle to obtain a suitable bound for $\left\|x(s)-x(t-)\right\|_E$. The only thing I'm able to deduce is that $\left\|x(s)-x(t_i-)\right\|_E<\frac\varepsilon3$ and $\left\|x(t)-x(t_i-)\right\|_E<\frac\varepsilon3$.

There are very easy counter-examples. Let $x(t)=1$ for $t \geq 0$ and $0$ for $t <0$. Let $t=\delta/2, s=-\delta/4$ and $u=\frac {3\delta} 4$. Then $|x(u)-x(s)-\Delta x(t)|=1$. [I am taking $E=\mathbb R$ and $K=[-1,1]$].

Note that $\sup_{-1\leq t \leq 1} |x(t)-x(t+h)| \geq |0-1|=1$ for any $h$ so uniform right continuity does not hold. Also, $\sup_{-1\leq t \leq 1} |x(t+)-x(s+)| \geq 1$ whenever $t>0$ and $s <0$ so $x(t+)$ is not uniformly continuous on $[-1,1]$. [By $x(t+)$ I mean the right hand limit of $x$ at $t$].